CBR has the answer

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18 years 7 months ago #14672 by Joe Keller
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It seems that extragalactic hydrogen atoms radiate at 2.80 K due to FitzGerald-Lorentz contraction of the spinning, orbiting electron. The mean free path of these photons needed, to redshift them to the CMB temperature, is confirmed, roughly, by the observed "convergence depth" of galactic, and CMB source, velocities. It is also confirmed, roughly, by Univ. of Washington theoretical calculations of the mean free path of extragalactic cosmic ray protons colliding with CMB photons; presumably CMB photons colliding with each other, have that same mean free path.

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18 years 7 months ago #17230 by Joe Keller
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The effective nuclear charge, Z, for helium, is 1.5 due to the presence of the fellow electron, so the effective "alpha"=Z*q^2/(hbar*c), for the helium Bohr model is 1.5x larger than for hydrogen. According to the theory above, the temperature of helium should be 1.5^4=5x that of the CMB (i.e., of hydrogen); this agrees well with the far CIRB peak.

The effective Z for oxygen, acting on its inner S1 electrons, is, because of "P2 orbital" electrons, only roughly 0.005 less than 7.5. Carbon, nitrogen and neon, also abundant in the universe, broaden that peak. For oxygen, 7.495^4*2.8016=8841 K, hotter than the near CIRB+cosmic optical background peak (which is downshifted by dust and by cosmological redshift), but equal to the surface temperature of typical bright stars. It is the inflection point of the Hertzsprung-Russell diagram (plotted on log-log paper)(see below).

For iron (neglecting P, D, and outer S orbital electrons), T=1.15 million degrees and the peak is 0.30 keV. The soft cosmic X-ray background lacks Planck shape and is said to range mainly from 0.5 to 2 keV or, according to another authority, from one to ten million degrees. Copper and zinc also fall in the lower end of the range.

For lead the peak is 44.1 keV, for thorium 69.9 keV and for uranium 78.2 keV. So, the heaviest stable elements match the (also somewhat irregular) hard CXB peak of 40 keV. For these heavy elements, use the relativistic formula 3*k*T=m*c^2*(gamma-1)*(1-gamma^(-2)), where gamma=sec(arcsin(beta)) and beta=(Z-0.5)*alpha.

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18 years 7 months ago #17005 by Ryan2006
Joe, I have been talking with Harry and he has been reading about the effects of plasma in the calculations of CBR have you included this in your calculations?

ryan Henningsgaard

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18 years 7 months ago #14690 by Joe Keller
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Hi Ryan!

Nope. What effects?

- Joe

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18 years 7 months ago #14693 by Ryan2006
According to this new calculation that includes plasma it distorts the redshift analysis.

ryan Henningsgaard

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18 years 7 months ago #17308 by Joe Keller
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Hi Ryan!

Thanks for the information!

- Joe Keller

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